A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f₁ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f₂ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of f₁ and f₂ will be :-

Both lenses are plano-convex, so the second surface is plane ($R = \infty$), which simplifies the lens maker’s formula:

$$\frac{1}{f} = \left( \frac{n}{n_m} - 1 \right) \left( \frac{1}{R} - 0 \right)$$

where:

• $n$ is the refractive index of the lens material,
• $n_m$ is the refractive index of the surrounding medium,
• $R$ is the radius of curvature of the curved surface.

Step 1: Calculate $f_1$ (Lens in Air)

For Lens 1 in Air:

$$\frac{1}{f_1} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{2} \right)$$

 $$\frac{1}{f_1} = \left( 1.5 - 1 \right) \times \frac{1}{2}$$ $$\frac{1}{f_1} = 0.5 \times \frac{1}{2} = \frac{0.5}{2} = \frac{1}{4}$$ $$f_1 = 4 \text{ cm}$$ 

Step 2: Calculate $f_2$ (Lens in Liquid)

For Lens 2 in Liquid:

$$\frac{1}{f_2} = \left( \frac{1.5}{1.2} - 1 \right) \left( \frac{1}{3} \right)$$

 $$\frac{1}{f_2} = \left( \frac{1.5 - 1.2}{1.2} \right) \times \frac{1}{3}$$ $$\frac{1}{f_2} = \left( \frac{0.3}{1.2} \right) \times \frac{1}{3}$$ $$\frac{1}{f_2} = \frac{0.3}{3.6}$$ $$f_2 = \frac{3.6}{0.3} = 12 \text{ cm}$$ 

Step 3: Find the Ratio $\frac{f_1}{f_2}$

$$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$$

Final Answer:

$$\boxed{\frac{1}{3}}$$